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I have to show that $$\int_{0}^{2\pi}\frac{d\theta}{(a^{2}\cos^{2}\theta+b^{2}\sin^{2}\theta)^{2}} =\frac{ \pi(a^{2}+b^{2})}{a^{3}b^{3}}$$ where $a,b>0$. I have tried using double angle formulas and Euler's trig identities to simplify this in order to use either residues or Cauchy's integral formula, but everything I have tried has made it messier and messier. Either I'm trying to simplify some things too early or something, but any help or tips would be much appreciated.

Okay, I made the substitutions, and right now just focusing on the denominator, I got to $$(a^{2}(z+z^{-1})^2-b^{2}(z-z^{-1})^2)^2$$ (i took out the common factor of 4 that I got). Then here is where I'm stuck. It looks like I have a difference of 2 perfect squares, so I then wrote $$((a(z+z^{-1})-b(z+z^{-1}))(a(z+z^{-1})+b(z+z^{-1})))^2$$ which I then wrote as $$(a(z+z^{-1})-b(z+z^{-1}))^2(a(z+z^{-1})+b(z+z^{-1}))^2$$ If I did it right, which I hope I did, Now I'm not sure where to go from here into getting a rational function.

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1 Answer

Just use the standard substitutions to turn this into an integral over the unit circle, i.e., $z=e^{i\theta}$, $d\theta = i e^{i\theta} dz$, $dz = \frac{d\theta}{iz}$, $\cos \theta = \frac{z+1/z}2$, and $\sin \theta = \frac{z-1/z}{2i}$. This will turn it into an integral of a rational function over the unit circle, then apply the residue theorem. If you get stuck somewhere on the way, update your question with exactly what you tried.

If $a^2=b^2$, then the integrand becomes the constant $1/a^4$, so the integral is $2\pi/a^4$, which agrees with the expression on the right for this case. Otherwise some admittedly painful calculations turn the integral into $$ \begin{align*} &\frac{16}{i(a^2-b^2)^2} \int_{|z|=1} \frac{z^3}{\left(z^4 + 2 \frac{a^2+b^2}{a^2-b^2} z^2 + 1\right)^2} \, dz \\ &= \frac{16}{i(a^2-b^2)^2} \int_{|z|=1} \frac{z^3} {\left(z^2+\frac{a+b}{a-b}\right)^2\left(z^2+\frac{a-b}{a+b}\right)^2}\, dz \end{align*} $$ Now we may assume $b>a>0$, so that the two poles in the unit disk are $z_{1,2} = \pm \sqrt{\frac{b-a}{b+a}}$. Each one of these is a double pole, so there are more painful calculations ahead that should lead you to the formula you are trying to prove.

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We cannot apply Cauchy's residue theorem because the poles lie on the unit circle. –  Américo Tavares Nov 18 '12 at 17:07
    
No, the poles will not be on the unit circle, why do you think so? –  Lukas Geyer Nov 18 '12 at 17:53
    
I've got $$\int_{C}-16i\frac{z^{3}}{\left( z^{2}+1\right) ^{4}\left( a^{2}-b^{2}\right) ^{2}}dz$$ –  Américo Tavares Nov 18 '12 at 19:18
    
+1. Sorry, my computation was wrong. There's a typo in $$\left( z^{4}+2\frac{a^{2}+b^{2}}{a^{2}-b^{2}}z^{2}+1\right) ^{2}=\left( z^{2}+\frac{a+b}{a-b}\right) ^{2}\left( z^{2}+\frac{a-b}{a+b}\right) ^{2}$$ –  Américo Tavares Nov 18 '12 at 22:46
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@AméricoTavares: Thanks for catching my mistake, fixed. –  Lukas Geyer Nov 19 '12 at 17:33
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